Problem: $\log_{\,81}{\dfrac{1}{27}}$ =
The equations $\log_b{y}=x$ and $b^x=y$ mean exactly the same thing according to the definition of logarithms. Therefore, we can rewrite our question as an exponential equation. Do remember that for the equations to be equivalent, we need $y$ and $b$ to be positive numbers, and $b\neq 1$. [Why can't b = 1?] So if $\log_{81}{\dfrac{1}{27}}=x$, then $81^x=\dfrac{1}{27}$. $81$ to what power is $\dfrac{1}{27}$ ? Notice that the fourth root of $81$ is $3$ and that $3^{-3}=\dfrac{1}{27}$. Since a m n = ( a √ n ) m a\^{\frac mn}=\left(\sqrt[ n]{a}\right)^m, it follows that ( 81 − − √ 4 ) − 3 = 81 − 3 4 = 1 27 \left(\sqrt[4]{81}\right)^{-3}=81\^{ -\frac34}=\dfrac{1}{27}. So $\log_{81}{\dfrac{1}{27}}=-\dfrac34$. Note: If answering the above question was difficult, you can solve the equation $81^x=\dfrac{1}{27}$ to get the answer. $\log_{81}{\dfrac{1}{27}}=-\dfrac34$